Let the plane $$ax + by + cz = d$$ pass through $$(2, 3, -5)$$ and is perpendicular to the planes $$2x + y - 5z = 10$$ and $$3x + 5y - 7z = 12$$.
If $$a, b, c, d$$ are integers $$d > 0$$ and $$\gcd(|a|, |b|, |c|, d) = 1$$ then the value of $$a + 7b + c + 20d$$ is equal to

We need to find the plane $$ax + by + cz = d$$ passing through $$(2, 3, -5)$$ and perpendicular to both $$2x + y - 5z = 10$$ and $$3x + 5y - 7z = 12$$.

Find the normal to the required plane:

The normal $$(a, b, c)$$ must be perpendicular to both $$(2, 1, -5)$$ and $$(3, 5, -7)$$.

$$\vec{n} = (2, 1, -5) \times (3, 5, -7) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -5 \\ 3 & 5 & -7 \end{vmatrix}$$

$$= \hat{i}(-7 + 25) - \hat{j}(-14 + 15) + \hat{k}(10 - 3)$$

$$= 18\hat{i} - \hat{j} + 7\hat{k}$$

So $$(a, b, c) = (18, -1, 7)$$.

Find $$d$$:

The plane passes through $$(2, 3, -5)$$:

$$18(2) + (-1)(3) + 7(-5) = 36 - 3 - 35 = -2$$

So $$d = -2$$.

Since we need $$d > 0$$, multiply through by $$-1$$:

$$(a, b, c, d) = (-18, 1, -7, 2)$$

Verify $$\gcd(|a|, |b|, |c|, d) = 1$$:

$$\gcd(18, 1, 7, 2) = 1$$ $$\checkmark$$

Compute $$a + 7b + c + 20d$$:

$$= -18 + 7(1) + (-7) + 20(2) = -18 + 7 - 7 + 40 = 22$$

The correct answer is Option D: $$\boxed{22}$$.